64bit - x64 instruction encoding (r/m, reg vs reg, r/m) -


what's difference in encoding (modrm:r/m, modrm:reg) vs (modrm:reg, modrm:r/m)? instruction cmpxchg vs divpd. thought register , address encoded in first byte , sib , displacement in second byte if needed? here's code:

    static void writeregistertomemory(icollection<byte> bytes, iregistertomemoryinstruction instruction, byte rex)     {         iaddress address = instruction.address;         byte register = instruction.register;          if (address.needsrex)         {             rex |= 0x40;             if (address.rexb)                 rex |= 1;             if (address.rexx)                 rex |= 1 << 1;         }          if (register > 7)             rex |= 0x44;        // rex.r         if (rex != 0)             bytes.add(rex);          bytes.addrange(instruction.opcode);         byte modrm = (byte)((register % 8) << 3);         modrm |= address.getmodrmaddressbyte();         bytes.add(modrm);         address.writescaledindexbyteanddisplacement(bytes);     }

so these 2 instructions encoded same different opcodes? (adds on page 457 of intel x64 manual)

op/en operand 1        operand 2  rm    modrm:reg (r, w) modrm:r/m (r)  mr    modrm:r/m (r, w) modrm:reg (r)

there isn't difference w.r.t. encoding, difference in 1 source , 1 destination. instructions have r/m source, except things cmpxchg, bts, xadd, xchg ambiguous (it's symmetric), alu ops have r/m, r form , r/m, imm form, , mov's memory. in encoding instructions (even if both operands registers), careful "which way around" are, or might end operands swapped. that's all, there in end no difference in how encoded.


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